YES

We show the termination of the TRS R:

  g(A()) -> A()
  g(B()) -> A()
  g(B()) -> B()
  g(C()) -> A()
  g(C()) -> B()
  g(C()) -> C()
  foldB(t,|0|()) -> t
  foldB(t,s(n)) -> f(foldB(t,n),B())
  foldC(t,|0|()) -> t
  foldC(t,s(n)) -> f(foldC(t,n),C())
  f(t,x) -> |f'|(t,g(x))
  |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
  |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
  |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
  |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)
  fold(t,x,|0|()) -> t
  fold(t,x,s(n)) -> f(fold(t,x,n),x)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: foldB#(t,s(n)) -> foldB#(t,n)
p3: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p4: foldC#(t,s(n)) -> foldC#(t,n)
p5: f#(t,x) -> |f'|#(t,g(x))
p6: f#(t,x) -> g#(x)
p7: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())
p8: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p9: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p10: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p11: fold#(t,x,s(n)) -> f#(fold(t,x,n),x)
p12: fold#(t,x,s(n)) -> fold#(t,x,n)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)
r16: fold(t,x,|0|()) -> t
r17: fold(t,x,s(n)) -> f(fold(t,x,n),x)

The estimated dependency graph contains the following SCCs:

  {p12}
  {p1, p2, p3, p4, p5, p7, p8, p9, p10}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: fold#(t,x,s(n)) -> fold#(t,x,n)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)
r16: fold(t,x,|0|()) -> t
r17: fold(t,x,s(n)) -> f(fold(t,x,n),x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          fold#_A(x1,x2,x3) = max{2, x1, x2, x3 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        fold# = s
      
      partial status:
    
        pi(fold#) = [1, 2, 3]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          fold#_A(x1,x2,x3) = max{x1 + 1, x2 + 1, x3 - 1}
          s_A(x1) = x1
    
      precedence:
      
        fold# = s
      
      partial status:
    
        pi(fold#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p4: foldB#(t,s(n)) -> foldB#(t,n)
p5: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p6: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p7: foldC#(t,s(n)) -> foldC#(t,n)
p8: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p9: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)
r16: fold(t,x,|0|()) -> t
r17: fold(t,x,s(n)) -> f(fold(t,x,n),x)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          foldB#_A(x1,x2) = max{x1 - 15, x2 + 37}
          s_A(x1) = max{173, x1 + 32}
          f#_A(x1,x2) = max{x1 - 21, x2 + 68}
          foldB_A(x1,x2) = max{x1, x2 + 57}
          B_A = 0
          |f'|#_A(x1,x2) = max{x1 - 21, x2 + 68}
          g_A(x1) = max{0, x1 - 1}
          triple_A(x1,x2,x3) = max{79, x2 + 67}
          A_A = 0
          |0|_A = 1
          |f''|#_A(x1) = max{62, x1 - 11}
          foldC#_A(x1,x2) = max{68, x1 - 18}
          foldC_A(x1,x2) = max{89, x1 + 2}
          C_A = 0
          |f''|_A(x1) = max{80, x1 + 10}
          |f'|_A(x1,x2) = max{x1, x2 + 89}
          f_A(x1,x2) = max{x1, x2 + 89}
    
      precedence:
      
        foldB > foldC = |f''| = |f'| = f > foldB# = f# = |f'|# = triple = |f''|# = foldC# > g > B > s = A = |0| = C
      
      partial status:
    
        pi(foldB#) = []
        pi(s) = []
        pi(f#) = []
        pi(foldB) = [1]
        pi(B) = []
        pi(|f'|#) = []
        pi(g) = []
        pi(triple) = []
        pi(A) = []
        pi(|0|) = []
        pi(|f''|#) = []
        pi(foldC#) = []
        pi(foldC) = []
        pi(C) = []
        pi(|f''|) = []
        pi(|f'|) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          foldB#_A(x1,x2) = 78
          s_A(x1) = 77
          f#_A(x1,x2) = 78
          foldB_A(x1,x2) = 76
          B_A = 68
          |f'|#_A(x1,x2) = 78
          g_A(x1) = 74
          triple_A(x1,x2,x3) = 76
          A_A = 67
          |0|_A = 66
          |f''|#_A(x1) = 78
          foldC#_A(x1,x2) = 78
          foldC_A(x1,x2) = 76
          C_A = 71
          |f''|_A(x1) = 76
          |f'|_A(x1,x2) = 76
          f_A(x1,x2) = 76
    
      precedence:
      
        foldB > B = g > C > foldB# = f# = |f'|# = A = |f''|# = foldC# > foldC = |f''| = |f'| = f > s > |0| > triple
      
      partial status:
    
        pi(foldB#) = []
        pi(s) = []
        pi(f#) = []
        pi(foldB) = []
        pi(B) = []
        pi(|f'|#) = []
        pi(g) = []
        pi(triple) = []
        pi(A) = []
        pi(|0|) = []
        pi(|f''|#) = []
        pi(foldC#) = []
        pi(foldC) = []
        pi(C) = []
        pi(|f''|) = []
        pi(|f'|) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(t,x) -> |f'|#(t,g(x))
p2: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p3: foldB#(t,s(n)) -> foldB#(t,n)
p4: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p5: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p6: foldC#(t,s(n)) -> foldC#(t,n)
p7: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p8: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)
r16: fold(t,x,|0|()) -> t
r17: fold(t,x,s(n)) -> f(fold(t,x,n),x)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p5, p6, p7, p8}
  {p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(t,x) -> |f'|#(t,g(x))
p2: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())
p3: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p4: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p5: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p6: foldC#(t,s(n)) -> foldC#(t,n)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)
r16: fold(t,x,|0|()) -> t
r17: fold(t,x,s(n)) -> f(fold(t,x,n),x)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (40,29)
            |f'|#_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (36,23)
            g_A(x1) = ((1,0),(1,0)) x1 + (3,1)
            triple_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (0,6)
            B_A() = (11,11)
            A_A() = (6,0)
            |f''|#_A(x1) = ((1,0),(1,1)) x1 + (39,2)
            foldB_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,18)
            s_A(x1) = ((1,0),(1,1)) x1 + (36,0)
            |0|_A() = (1,7)
            foldC#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (37,4)
            foldC_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,41)
            C_A() = (9,1)
            |f''|_A(x1) = x1 + (3,42)
            |f'|_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (36,2)
            f_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (36,6)
    
      precedence:
      
        |0| = foldC > f# = |f'|# = |f''|# > foldB = foldC# > s = |f'| = f > g = C > A > triple = B = |f''|
      
      partial status:
    
        pi(f#) = []
        pi(|f'|#) = [2]
        pi(g) = []
        pi(triple) = []
        pi(B) = []
        pi(A) = []
        pi(|f''|#) = []
        pi(foldB) = []
        pi(s) = [1]
        pi(|0|) = []
        pi(foldC#) = []
        pi(foldC) = []
        pi(C) = []
        pi(|f''|) = [1]
        pi(|f'|) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = (10,17)
            |f'|#_A(x1,x2) = ((1,0),(1,1)) x2 + (1,1)
            g_A(x1) = (8,7)
            triple_A(x1,x2,x3) = (3,4)
            B_A() = (11,6)
            A_A() = (6,5)
            |f''|#_A(x1) = (5,6)
            foldB_A(x1,x2) = (5,1)
            s_A(x1) = (2,0)
            |0|_A() = (2,0)
            foldC#_A(x1,x2) = (1,0)
            foldC_A(x1,x2) = (8,2)
            C_A() = (7,1)
            |f''|_A(x1) = x1 + (0,1)
            |f'|_A(x1,x2) = (4,3)
            f_A(x1,x2) = (4,3)
    
      precedence:
      
        foldB > f# = foldC > |f'|# = g = B = |0| = |f'| = f > A = |f''| > triple > |f''|# = s = foldC# = C
      
      partial status:
    
        pi(f#) = []
        pi(|f'|#) = [2]
        pi(g) = []
        pi(triple) = []
        pi(B) = []
        pi(A) = []
        pi(|f''|#) = []
        pi(foldB) = []
        pi(s) = []
        pi(|0|) = []
        pi(foldC#) = []
        pi(foldC) = []
        pi(C) = []
        pi(|f''|) = []
        pi(|f'|) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p5

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p2: foldC#(t,s(n)) -> foldC#(t,n)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)
r16: fold(t,x,|0|()) -> t
r17: fold(t,x,s(n)) -> f(fold(t,x,n),x)

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: foldC#(t,s(n)) -> foldC#(t,n)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)
r16: fold(t,x,|0|()) -> t
r17: fold(t,x,s(n)) -> f(fold(t,x,n),x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          foldC#_A(x1,x2) = max{2, x1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        foldC# = s
      
      partial status:
    
        pi(foldC#) = [1, 2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          foldC#_A(x1,x2) = max{x1, x2 - 1}
          s_A(x1) = x1
    
      precedence:
      
        foldC# = s
      
      partial status:
    
        pi(foldC#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: foldB#(t,s(n)) -> foldB#(t,n)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)
r16: fold(t,x,|0|()) -> t
r17: fold(t,x,s(n)) -> f(fold(t,x,n),x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          foldB#_A(x1,x2) = max{2, x1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        foldB# = s
      
      partial status:
    
        pi(foldB#) = [1, 2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          foldB#_A(x1,x2) = max{x1, x2 - 1}
          s_A(x1) = x1
    
      precedence:
      
        foldB# = s
      
      partial status:
    
        pi(foldB#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.